Question
Sun August 30, 2009 By: Suchet Raghavan

triangles

Expert Reply
Mon August 31, 2009

Pythagoras' Theorem,

In ΔPQR

PR2 = PQ2 + QR2

QR = 2QS

PR2 = PQ2 + (2QS)2

PR2 = PQ2 + 4QS2 ..........(1)

In ΔPQS

PS2 = PQ2 + QS2

QS2 = PS2 - PQ2

 Using this in (1),

PR2 = PQ2 + 4(PS2 - PQ2)   

PR2 = 4PS2 - 3PQ2

Regards,

Team,

TopperLearning. 

Home Work Help